(a) Show that the number of photons per unit volume in a photon gas of temperature $\mathbf{T}$is approximately$\mathbf{(}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{3}}\mathbf{}{\mathbf{\text{m}}}^{\mathbf{-}\mathbf{3}}\mathbf{)}{\mathbf{T}}^{\mathbf{3}}\mathbf{\cdot }$(Note:${\int }_0^{\infty }{x}^2{(e^x-1)}^{-1}\mathrm{dx}\cong 2.40.)$

(b)Combine this with a result derived in Example 9.6 to show that the average photon energy in a cavity at temperature$\mathbf{T}$is given by$\overline{\mathbf{E}}\mathbf{\cong }\mathbf{2}\mathbf{.7}{\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}$.

General equation for$\mathbf{N}$particles of E energy is given by

role="math" localid="1660030263876" $\mathbf{N}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{N}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{dE}$…..(1)

Where,

$\mathbf{N}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{\to }$Quantum distribution

$\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{\to }$ Density of states

$\mathbf{N}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{e}}^{\frac{\mathbf{E}}{{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}\mathbf{-}\mathbf{1}}$ ……(2)

Where,${\mathbf{K}}_{\mathbf{B}}$Boltzmann's constant.

Also,

$\mathbf{D}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{8}\mathbf{\pi V}}{{\mathbf{h}}^{\mathbf{3}}{\mathbf{c}}^{\mathbf{3}}}{\mathbf{E}}^{\mathbf{2}}$ ……(3)

Where,

$\mathbf{V}\mathbf{\to }$Volume of container

$\mathbf{h}\mathbf{\to }$being Planck's constant

$\mathbf{c}\mathbf{\to }$Speed of light in vacuum.

Expression for energy of photons in a container is given by-

$\mathbf{U}\mathbf{=}\frac{\mathbf{8}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{Vk}}_{\mathbf{B}}^{\mathbf{4}}{\mathbf{T}}^{\mathbf{4}}}{\mathbf{15}{\mathbf{h}}^{\mathbf{3}}{\mathbf{c}}^{\mathbf{3}}}$…..(4)

Where,

${\mathbf{k}}_{\mathbf{B}}\mathbf{\to }$Boltzmann's constant

$\mathbf{h}\mathbf{\to }$Planck's constant

$\mathbf{c}\mathbf{\to }$Speed of light in vacuum

$\mathbf{T}\mathbf{\to }$Temperature

$\mathbf{V}\mathbf{\to }$Volume of container

To get an expression for the number of photon per unit volume at some temperature R , equations (2) and (3) are inserted in equation (1), and simplified:

$\begin{array}{c}\mathrm{N}={\int }_0^{\mathrm{\infty }}\mathrm{N}\left(\mathrm{E}\right)\mathrm{D}\left(\mathrm{E}\right)\mathrm{dE}\\ ={\int }_0^{\mathrm{\infty }}\left[\frac{1}{{\mathrm{e}}^{\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}-1}\right]\left[\frac{8\mathrm{\pi V}}{{\mathrm{\hslash }}^3{\mathrm{c}}^3}{\mathrm{E}}^2\right]\mathrm{dEN}\\ =\frac{8\mathrm{\pi V}}{{\mathrm{\hslash }}^3{\mathrm{c}}^3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{E}}{{\mathrm{e}}^{\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}-1}\mathrm{dE}\end{array}$

That can be simplified more by the use of the substitutions:

$\begin{array}{c}\mathrm{x}=\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\\ \mathrm{E}={\mathrm{xk}}_{\mathrm{B}}\mathrm{T}\\ \mathrm{dE}={\mathrm{k}}_{\mathrm{B}}\mathrm{Tdx}\end{array}$

So then substitute those into equation (6)

$\begin{array}{c}\mathrm{N}=\frac{8\mathrm{\pi V}}{{\mathrm{h}}^3{\mathrm{c}}^3}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{E}}{{\mathrm{e}}^{\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}-1}\mathrm{dE}\\ \mathrm{N}=\frac{8\mathrm{\pi V}}{{\mathrm{h}}^3{\mathrm{c}}^3}{\int }_0^{\mathrm{\infty }}\frac{{\left({\mathrm{xk}}_{\mathrm{B}}\mathrm{T}\right)}^2}{{\mathrm{e}}^{\mathrm{x}}-1}\left({\mathrm{k}}_{\mathrm{B}}\mathrm{Tdx}\right)\end{array}$

$\mathrm{N}=\frac{8{\mathrm{\pi Vk}}_{\mathrm{B}}^3{\mathrm{T}}^3}{{\mathrm{h}}^3{\mathrm{c}}^3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}$

Equation (5) can then be used to fill in for the integral, $\mathrm{N}=\frac{8{\mathrm{\pi Vk}}_{\mathrm{B}}^3{\mathrm{T}}^3}{{\mathrm{\hslash }}^3{\mathrm{c}}^3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}$

Substitute $1.38\times {10}^{23}\text{J}/\text{K}$for ${\mathrm{k}}_{\mathrm{B}},6.63\times {10}^{-34}\text{J}.\text{s}$ for $\mathrm{\hslash }$, and $3\times {10}^8\text{m}/\text{s}$ for c

$\begin{array}{l}\mathrm{N}=\frac{8\mathrm{\pi V}{(1.38\times {10}^{-23}\text{J}/\text{K})}^3{\mathrm{T}}^3}{{(6.63\times {10}^{-34}\text{J}.\text{s})}^3{(3\times {10}^8\text{m}/\text{s})}^3}(2.40)\\ \mathrm{N}=\left(2\times {10}^7\frac{1}{{\mathrm{m}}^3.{\mathrm{K}}^3}\right){\mathrm{VT}}^3\end{array}$

To find the average energy of a photon isa cavity of volume $\mathrm{V}$, use that the number of photons$\mathrm{N}$times the average energy of the photons$E$is equal to the total energy of the photons$\mathrm{U}$:

$\begin{array}{l}\mathrm{U}=\mathrm{NE}\\ \mathrm{E}=\frac{\mathrm{U}}{\mathrm{N}}\end{array}$

Put the value and solve

$\begin{array}{c}\mathrm{E}=\frac{\left(\frac{8{\mathrm{\pi }}^5{\mathrm{v}}_{\mathrm{B}}^4{\mathrm{T}}^4}{15{\mathrm{h}}^3{\mathrm{c}}^3}\right)}{\left(\frac{8{\mathrm{\pi Vk}}_{\mathrm{B}}^3{\mathrm{T}}^3}{{\mathrm{\hslash }}^3{\mathrm{c}}^3}\right){\displaystyle {\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}-1}}\mathrm{dx}}\\ \mathrm{E}=\frac{{\mathrm{\pi }}^4{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{15{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}-1}}\mathrm{dx}}\end{array}$

That can be simplified with equation (5) again

$\begin{array}{l}\mathrm{E}=\frac{{\mathrm{\pi }}^4{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{15{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2}{{\mathrm{e}}^{\mathrm{x}}-1}}\mathrm{dx}}\mathrm{n}\\ \mathrm{E}=\frac{{\mathrm{\pi }}^4}{15\left(2.40\right)}{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\\ \mathrm{E}\cong 2.7{\mathrm{k}}_{\mathrm{B}}\mathrm{T}\end{array}$